## Precalculus (6th Edition) Blitzer

The product of $AB={{I}_{2}}$, product of $BA={{I}_{2}}$ and $B$ is the multiplicative inverse of $A$: $B={{A}^{-1}}$.
The provided expression is $A=\left[ \begin{matrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ \end{matrix} \right]$ Now, we will compute the matrix as $\left[ AB \right]$ \begin{align} & AB=\left[ \begin{matrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \left( -2 \right)\times 1+1\times 3 & \left( -2 \right)\times 2+1\times 4 \\ \frac{3}{2}\times 1+\left( -\frac{1}{2} \right)\times 3 & \frac{3}{2}\times 2+\left( -\frac{1}{2} \right)\times 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{2}} \end{align} Now, we will compute the matrix as $\left[ BA \right]$ \begin{align} & BA=\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\times \left( -2 \right)+2\times \frac{3}{2} & 1\times 1+2\times \left( -\frac{1}{2} \right) \\ 3\times \left( -2 \right)+4\times \frac{3}{2} & 3\times 1+4\times \left( -\frac{1}{2} \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{2}} \end{align} Both matrix $AB$ and matrix $BA$ are equal to the identity matrix Now, we will compute the matrix $B={{A}^{-1}}$. $B=\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ \end{matrix} \right]$ And $A=\left[ \begin{matrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \\ \end{matrix} \right]$ Determine the inverse of matrix A. Now, using inverse formula ${{A}^{-1}}=\frac{1}{\left| ad-bc \right|}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$ Substitute the values to get \begin{align} & a=-2 \\ & b=1 \\ & c=\frac{3}{2} \\ & d=-\frac{1}{2} \\ \end{align} So, \begin{align} & {{A}^{-1}}=\frac{1}{\left| 1-\frac{3}{2} \right|}\left[ \begin{matrix} -\frac{1}{2} & -1 \\ -\frac{3}{2} & -2 \\ \end{matrix} \right] \\ & =\frac{2}{-1}\left[ \begin{matrix} -\frac{1}{2} & -1 \\ -\frac{3}{2} & -2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ \end{matrix} \right] \end{align} Thus, $B$ is the multiplicative inverse of $A$.